Daily math problems for October 14, 2026 — Euclidean Geometry, Geometry, Calculus & more

$A=5\times 3=15{\text{ cm}}^2$.

The line is in slope-intercept form $y=mx+b$ with $m=-3$ and $b=5$.

So the slope is $-3$.

$f^{\mathrm{\prime }}(x)=2x-6$.

Setting $f^{\mathrm{\prime }}(c)=0$: $2c-6=0⟹c=3$.

Since $f^{\mathrm{\prime }\mathrm{\prime }}(x)=2>0$, this is indeed a minimum.

The number of doubling periods is $$\frac{12}{3}=4.$$ Each period multiplies the population by $2$, so $$N=500\cdot 2^4=500\cdot 16=8000.$$

Multiply along each branch to get the joint probabilities: $$P(A\cap D)=0.6\times 0.05=0.03$$ $$P(B\cap D)=0.4\times 0.10=0.04$$ The total probability of a defective item is $$P(D)=0.03+0.04=0.07$$ Apply the conditional probability formula: $$P(A\mid D)=\frac{0.03}{0.07}=\frac{3}{7}$$

Negation of $\mathrm{\forall }$ is $\mathrm{\exists }$; negation of $>$ is $\le$. So the negation is $\mathrm{\exists }x\in \mathbb{R}:x^2+1\le 0$. (This statement is itself false, but the question asked for the negation, not its truth value.)

$\bullet$ Compute $\det A$ by cofactor expansion along row $1$:

$$\det A=0\cdot M_{11}-(-1)\cdot M_{12}+3\cdot M_{13}=2+3=5$$

$\bullet$ For an $n\times n$ matrix, $\det (cM)=c^n\det M$ and $\det (M^{-1})=1\mathrm{/}\det M$:

$$\det (5A^{-1})=5^3\cdot \frac{1}{\det A}=\frac{125}{5}=25$$

$g(0)=1>0$, $g(\pi \mathrm{/}2)=0-\pi \mathrm{/}2<0$. Since $g$ is continuous, by the IVT there is $c\in (0,\pi \mathrm{/}2)$ with $g(c)=0$, i.e. $\cos c=c$.

$A(x)=10x-x^2$. $A^{\mathrm{\prime }}(x)=10-2x=0\Rightarrow x=5$. The maximum-area rectangle is the square $5\times 5$, with area $25$.

By Fermat, $3^6\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}7)$, so ${\widehat{3}}^6=\widehat{1}$.